2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.
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- A parenting magazine reports that the average amount of wireless data used by teenagers each month is 10 Gb. For her science fair project, Ella sets out to prove the magazine wrong. She claims that the mean among teenagers in her area is less than reported. Ella collects information from a simple random sample of 4 teenagers at her high school, and calculates a mean of 8.4 Gb per month with a standard deviation of 2.2 Gb per month. Assume that the population distribution is approximately normal. Test Ella's claim at the 0.05 level of significance. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Η :μ H₂H = 10 10Two friends, Karen and Jodi, work different shifts for the same ambulance service. They wonder if the different shifts average different numbers of calls. Looking at past records, Karen determines from a random sample of 41 shifts that she had a mean of 5.1 calls per shift. She knows that the population standard deviation for her shift is 1.5 calls. Jodi calculates from a random sample of 32 shifts that her mean was 5.8 calls per shift. She knows that the population standard deviation for her shift is 1.2 calls. Test the claim that there is a difference between the mean numbers of calls for the two shifts at the 0.01 level of significance. Let Karen's shifts be Population 1 and let Jodi's shifts be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.A professor believes that, for the introductory art history classes at his university, the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. He collects data from a random sample of 150 students in evening classes and finds that they have a mean test score of 88.8. He knows the population standard deviation for the evening classes to be 8.4 points. A random sample of 250 students from morning classes results in a mean test score of 89.9. He knows the population standard deviation for the morning classes to be 5.4 points. Test his claim with a 99% level of confidence. Let students in the evening classes be Population 1 and let students in the morning classes be Population 2. Step 3 of 3 : Draw a conclusion and interpret the decision.
- A parenting magazine reports that the average amount of wireless data used by teenagers each month is 10 Gb. For her science fair project, Ella sets out to prove the magazine wrong. She claims that the mean among teenagers in her area is less than reported. Ella collects information from a simple random sample of 4 teenagers at her high school, and calculates a mean of 8.4 Gb per month with a standard deviation of 2.2 Gb per month. Assume that the population distribution is approximately normal. Test Ella's claim at the 0.05 level of significance. Step 3 of 3: Draw a conclusion and interpret the decision. Answer Tables Keypad Keyboard Shortcuts We reject the null hypothesis and conclude that there is sufficient evidence at a 0.05 level of significance that the average amount of wireless data used by teenagers each month is less than 10 Gb. We reject the null hypothesis and conclude that there is insufficient evidence at a 0.05 level of significance that the average amount of wireless…A professor is concerned that the two sections of college algebra that he teaches are not performing at the same level. To test his claim, he looks at the mean exam score for a random sample of students from each of his classes. In Class 1, the mean exam score for 1515 students is 81.281.2 with a standard deviation of 3.43.4. In Class 2, the mean exam score for 1818 students is 83.883.8 with a standard deviation of 5.95.9. Test the professor’s claim at the 0.100.10 level of significance. Assume that both populations are approximately normal and that the population variances are equal. Let Class 1 be Population 1 and let Class 2 be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to three decimal places.A professor believes that, for the introductory art history classes at his university, the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. He collects data from a random sample of 250 students in evening classes and finds that they have a mean test score of 76.8. He knows the population standard deviation for the evening classes to be 7.2 points. A random sample of 200 students from morning classes results in a mean test score of 77.8. He knows the population standard deviation for the morning classes to be 1.9 points. Test his claim with a 90% level of confidence. Let students in the evening classes be Population 1 and let students in the morning classes be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.
- A Statistics instructor wanted to compare the scores on the same final exam given to students at a local community college and local university. The Statistics instructor feels that students at the community college would be better prepared and have higher exam scores on the final exam. After grading the final exams from both institutions, the Statistics instructor found that for a sample of 36 community college students, the sample had a mean score of 84 with a standard deviation of 3.4. A sample of 45 university students had a mean score 79 with a standard deviation of 2.6. Construct a 95% confidence interval for the difference between population mean exam scores. Round your final results to one decimal place. _____________ ≤ ??? − ????? ≤ ______________You wish to evaluate the differences in educational quality between two training centres (A and B). You take a random sample of 98 individuals trained at centre A, and an independent random sample of 41 individuals trained at centre B. The individuals in both samples take the same, standardised test. The mean test score of individuals in sample A is 85.1 and the sample standard deviation is 12.2. The mean test score in sample B is 67.4 and the sample standard deviation is 11.0. Find the lower limit of a 98% confidence interval for the difference between the population means. Form the confidence interval around a positive point estimate for the difference between the means. In order to obtain a positive point estimate, make sure you subtract the lower value from the higher value when calculating the difference between the sample means. (Provide your answer as a number rounded to 2 decimal places.)Two friends, Karen and Jodi, work different shifts for the same ambulance service. They wonder if the different shifts average different numbers of calls. Looking at past records, Karen determines from a random sample of 39 shifts that she had a mean of 4.7 calls per shift. She knows that the population standard deviation for her shift is 1.5 calls. Jodi calculates from a random sample of 31 shifts that her mean was 5.4 calls per shift. She knows that the population standard deviation for her shift is 1.2 calls. Test the claim that there is a difference between the mean numbers of calls for the two shifts at the 0.02 level of significance. Let Karen's shifts be Population 1 and let Jodi's shifts be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.
- A researcher thinks that people under forty have vocabularies different from those over sixty years of age. The researcher administers a vocabulary test to a group of 50 younger subjects and a group of 50 older subjects. Higher scores reflect better performance. The mean score for younger subjects was 9.0, and the standard deviation of younger subject's scores was 5.0. The mean score for older subjects was 20.0, and the standard deviation of older subject's scores was 8.0. Does this experiment provide evidence for the researcher's theory? As part of your answer: a. Please provide, in words, the null and alternative hypotheses (NO symbols). Note that you can choose whether to perform a one- or two-tailed test Answer: Null hypothesis: Alernative hypothesis: b. Provide a brief (one or two sentence(s) rationale for your decision to use a one or two tailed test Answer c. Calculate the Standard Error (SE) of the mean difference (demonstrate calculation steps to get an entire point) and…A professor believes that, for the introductory art history classes at his university, the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. He collects data from a random sample of 250 students in evening classes and finds that they have a mean test score of 88.1. He knows the population standard deviation for the evening classes to be 4.2 points. A random sample of 150 students from morning classes results in a mean test score of 89.1. He knows the population standard deviation for the morning classes to be 7.7 points. Test his claim with a 95 % level of confidence. Let students in the evening classes be Population 1 and let students in the morning classes be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to two decimal places.According to the U.S. Census, the average adult woman is the United States is 65 inches tall and the standard deviation is 3 inches. If Zsike is 67 inches tall, what is her z-score?
